The endomorphism ring is a field

[mathmari]

Hey!

Let $R$ be a commutative ring with unit and $M$ be a $R$-module.

I want to show that the endomorphism ring $\text{End}_R(M)=\text{Hom}_R(M,M)$ of a simple $R$-module is a field. We have that $\text{End}_R(M)=\text{Hom}_R(M,M)=\{f:M\rightarrow M \mid f \ : \ R-\text{ homomorphism}\}$.

We have that since $M$ s simple, it is cyclic and isomorphic to $R/J$, where $J$ is a maximal ideal of $R$.

So, to show that the endomorphism ring is a field do we have to show that the mapping $R\rightarrow \text{End}_R(R/J)$ is an homomorphism with kernel $J$ ? (Wondering)

 

[Euge]

Hi mathmari,

In http://mathhelpboards.com/linear-abstract-algebra-14/show-966-isomorphism-18576.html, you proved that if $R$ is a commutative ring with unity, then every nonzero homomorphism of simple $R$-modules is an isomorphism. Using that theorem, you can claim that every nonzero element of $\operatorname{End}_R(M)$ is an automorphism of $M$, and therefore invertible. Consequently, $\operatorname{End}_R(M)$ is a field.

 

[mathmari]

In http://mathhelpboards.com/linear-abstract-algebra-14/show-966-isomorphism-18576.html, you proved that if $R$ is a commutative ring with unity, then every nonzero homomorphism of simple $R$-modules is an isomorphism. Using that theorem, you can claim that every nonzero element of $\operatorname{End}_R(M)$ is an automorphism of $M$, and therefore invertible. Consequently, $\operatorname{End}_R(M)$ is a field.

How do we know that there are non-zero homomorphisms? (Wondering)

Also, how do we know that there are invertible elements, when we know that every nonzero element of $\operatorname{End}_R(M)$ is an automorphism of $M$ ? (Wondering)

 

[Euge]

How do we know that there are non-zero homomorphisms? (Wondering)

There may not be, in which case $\operatorname{End}_R(M)$ is the zero ring, which is a field.

Also, how do we know that there are invertible elements, when we know that every nonzero element of $\operatorname{End}_R(M)$ is an automorphism of $M$ ? (Wondering)

Automorphisms of $M$ are bijections, and bijections have inverses.